Direct sum of 2 subspaces to obtain $\mathbb{R}^4$ 
Consider $V$ as a subspace in $\mathbb{R}^4$ where $V= \text{span}\{(2,1,0,1),(2,-1,-1,-1),(3,0,2,3)\}$. Find a subspace $W$ so that  $\mathbb{R}^4=W\oplus V$.

If I find a vector in $W$ that is linearly independent of those in $V$ and it spans $W$, is that  enough for what i have to find?
 A: That is almost enough; you just also have to prove that the generators of $V$, $(2,1,0,1)$, $(2,-1,-1,-1)$, $(3,0,2,3)$, are linearly independent.
In short, you need find a vector $v$ and show that
$(2,1,0,1), (2,-1,-1,-1), (3,0,2,3),$ and $v$ are all linearly independent.  Then let $W = \text{span}(v)$.  This will imply (i) that $W \cap V = \{\textbf{0}\}$, and (ii) that $W + V = \mathbb{R}^4$.
A: Step 1. We use Gauss Elimination method to find a basis of V.
$$A=\left[ {\begin{array}{*{20}{c}}
  2&1&0&1 \\ 
  2&{ - 1}&{ - 1}&{ - 1} \\ 
  3&0&2&3 
\end{array}} \right]\xrightarrow[{2R_3 - 2R_1 \to R_3}]{{R_2 - R_1 \to R_2}}\left[ {\begin{array}{*{20}{l}}
  2&1&0&1 \\ 
  0&{ - 2}&{ - 1}&{ - 2} \\ 
  0&{ - 2}&4&4 
\end{array}} \right]\xrightarrow{{R_3 - R_2 \to R_3}}\left[ {\begin{array}{*{20}{c}}
  2&1&0&1 \\ 
  0&{ - 2}&{ - 1}&{ - 2} \\ 
  0&0&5&6 
\end{array}} \right]$$
Therefore, $V$ has a basis $\{ (2,1,0,1),(0,-2,-1,-2),(0,0,5,6)\}$. 
Step 2. Since the basis of $\mathbb{R}^4$ has four vectors. So we need to find 1 vector $u$ such that $u$ and these three vectors are linearly independent. We can choose $u=(0,0,0,1)$. Choose $W=span\{u\}$. Then $\mathbb{R}^4=W\oplus V$.
